14. Sum of Degrees of Vertices. The sum of the degrees of the vertices in a graph
with a finite
number of edges is twice the number of edges.
15. Connected. A graph is called connected if there is a path between each two
vertices of the
graph. We say two
8. Variance and Standard Deviation for Bernoulli Trials. In Bernoulli trials with
probability
p of success, the variance for one trial is p(1p) and for n trials is np(1p), so the
standard
deviation for n trials is !np(1 p).
9. Central Limit Theorem. The c
5. For which values of n does the complete graph on n vertices have an Eulerian
Circuit?
6. The hypercube graph Qn has as its vertex set the n-tuples of zeros and ones. Two
of these
vertices are adjacent if and only if they are different in one position.
If you analyze our proof of Diracs theorem, you will see that we really used only a
consequence
of the condition that all vertices have degree at least v/2, namely that for any two
vertices, the
sum of their degrees is at least n.
Theorem 6.12 (Ore) If G
been marked with an a and so is in Y A. But every other edge must be covered by
XA
because in a bipartite graph, each edge must be incident with a vertex in each part.
Therefore
C is a vertex cover. If an element of Y A, were not matched, it would be an
e
Tour of the connected component of G_ containing xk+1 to our tour. When we add the
last edge
and vertex of our closed path to the path we have been constructing, every vertex
and edge of the
graph will have to be in the path we have constructed, because e
listing all vertices adjacent to it. In the case of multiple edges, we list each
adjacency as many
times as there are edges that give the adjacency. In our pseudocode we implement
the idea of
an adjacency list with the array E that gives in position i a l
than the number of registers our computer has available. The problem of assigning
variables to
registers is called the register assignment problem.
An intersection graph of a set of intervals of real numbers is called an interval
graph. The
assignment of
nonempty children. Thus removing the root gives us two binary trees, rooted at the
children of
the original root, each with fewer than n vertices. By the definition of full, each of
the subtrees
rooted in the two children must be full binary tree. The num
measure of the size of the input graph, and k is independent of n), including
counting as one step
solving an instance of the first problem, and accepts exactly the instances of the
second problem
that have a yes answer. The question of whether a graph ha
components of the graph GW that results from removing the closed walk, and then
follow our
closed walk, pausing each time we enter a new connected component of G W to
recursively
construct an Eulerian Tour of the component and traverse it before returning
1
3
3
1
3
1
3
11
1
3
3
3
3
1
3
which vertex b is inside the cycle C and one in which it is outside C. (Notice also
that in both
cases, we have more than one choice for the cycle because there are two ways in
which we could
use the quadrilateral at the bot
5. Find all induced cycles in the graph of Figure 6.9.
6. What is the size of the largest induced Kn in Figure 6.9?
7. Find the largest induced Kn (in words, the largest complete subgraph) you can in
Figure
6.8.
8. Find the size of the largest induced Pn
vertex we add to the tree from vertex V [Vertex[m] will have distance n+1 from the
tree. Thus
every vertex added to the tree from a vertex of distance n from V [x] will have
distance n + 1
from V [x]. Further, all vertices of distance n + 1 are adjacent t
of connected components wont change. If the endpoints of the edge are in different
connected
components, then the number of connected components can go down by one. Since
an edge has
two endpoints, it is impossible for the number of connected components t
maximum matching, that is, a matching that is at least as big as any other
matching.
The graph in Figure 6.22 is an example of a bipartite graph. A graph is called
bipartite
whenever its vertex set can be partitioned into two sets X and Y so that each edg
not in S. Thus when it stops, S will be the vertex set of a connected component of
the graph
and E_ will be the edge set of a spanning tree of this connected component. This
suggests that
one use that we might make of algorithm Spantree is to find connect
NP-complete, 295, 296
NP-complete Problems, 294
number theory, 4081
one-to-one function, 11
one-way function, 68, 70
only if (in logic), 93
onto function, 11
or
exclusive (in logic), 85
or (in logic), 85, 86, 92
exclusive, 86
order
lexicographic, 13
order
polynomial time and the class NP of problems said to be solvable in
nondeterministic polynomial
time. We are not going to describe these problem classes in their full generality. A
course in formal
languages or perhaps algorithms is a more appropriate pla
trials large enough, we can make the probability that the number X of successes is
between
np ns and np + ns as close to 1 as we choose. For example, we can make the
probability
that the number of successes is within 1% (or 0.1 per cent) of the expected n
collision, 187, 191
collisions, 228, 234
empty slots, 228, 234
expected maximum number of keys per
slot, 233, 234
expected number of collisions, 228, 234
expected number of hashes until all slots
occupied, 230, 234
expected number of items per slot, 227,
13. Prove that a graph is an r-tree, as defined at the end of the section if and only if
it is a
rooted tree.
14. Use the inductive definition of a rooted tree (r-tree) given at the end of the
section to prove
once again that a rooted tree with n vertices
suitable. If you can do so, give an assignment of people to jobs. If you cannot, try to
explain why not.
Exercise 6.4-2 Table 6.2 shows a second sample of the kinds of applications a
school
district might get for its positions. Draw a graph as before and
3. A path that includes each vertex of the graph at least once and each edge of the
graph
exactly once, but has different first and last endpoints, is known as an Eulerian Trail
4. A graph G has an Eulerian Trail if and only if G is connected and all but
guarantee a Hamiltonian cycle, because when we got to the second last vertex on
the cycle, all
of the n 2 vertices it is adjacent to might already be on the cycle and different
from the first
vertex, so we would not have an edge on which we could leave th
sure that we do not examine edges that point from vertex i back to a vertex that is
already in
the tree.
This algorithm requires that we execute the For loop that starts in Line 4 once for
each
edge incident with vertex i. The While loop that starts in Li
15. Draw a graph of the equation y = x(1 x) for x between 0 and 1. What is the
maximum
value of y? Why does this show that the variance (see Problem 10 in this section) of
the
number of successes random variable for n independent trials is less than or
eq
measure of the size of the input graph, and k is independent of n), including
counting as
one step solving an instance of the first problem, and accepts exactly the instances
of the
second problem that have a yes answer.
298 CHAPTER 6. GRAPHS
Figure 6.19: